Problem: Recently Duff has been a soldier in the army. Malek is her commander.

Their country, Andarz Gu has n cities (numbered from 1 to n) and n - 1 bidirectional roads. Each road connects two different cities. There exist a unique path between any two cities.

There are also m people living in Andarz Gu (numbered from 1 to m). Each person has and ID number. ID number of i - th person is i and he/she lives in city number ci. Note that there may be more than one person in a city, also there may be no people living in the city.

<image>

Malek loves to order. That's why he asks Duff to answer to q queries. In each query, he gives her numbers v, u and a.

To answer a query:

Assume there are x people living in the cities lying on the path from city v to city u. Assume these people's IDs are p1, p2, ..., px in increasing order. 

If k = min(x, a), then Duff should tell Malek numbers k, p1, p2, ..., pk in this order. In the other words, Malek wants to know a minimums on that path (or less, if there are less than a people).

Duff is very busy at the moment, so she asked you to help her and answer the queries.

Input

The first line of input contains three integers, n, m and q (1 ≤ n, m, q ≤ 105).

The next n - 1 lines contain the roads. Each line contains two integers v and u, endpoints of a road (1 ≤ v, u ≤ n, v ≠ u).

Next line contains m integers c1, c2, ..., cm separated by spaces (1 ≤ ci ≤ n for each 1 ≤ i ≤ m).

Next q lines contain the queries. Each of them contains three integers, v, u and a (1 ≤ v, u ≤ n and 1 ≤ a ≤ 10).

Output

For each query, print numbers k, p1, p2, ..., pk separated by spaces in one line.

Examples

Input

5 4 5
1 3
1 2
1 4
4 5
2 1 4 3
4 5 6
1 5 2
5 5 10
2 3 3
5 3 1


Output

1 3
2 2 3
0
3 1 2 4
1 2

Note

Graph of Andarz Gu in the sample case is as follows (ID of people in each city are written next to them):

<image>

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
int n, m, q, h[100005], par[100005][20];
vector<int> ke[100005];
struct node {
  int a[11];
  void reset() {
    for (long long i = (0); i <= (10); i++) a[i] = 1000000007;
  }
  void add(int val) {
    a[10] = val;
    sort(a, a + 11);
  }
} val[100005], gt[100005][20];
bool giong(node p, node q) {
  for (long long i = (0); i <= (10); i++)
    if (p.a[i] != q.a[i]) return 0;
  return 1;
}
node addnode(node x, node y) {
  node res = x;
  for (long long i = (0); i <= (10); i++)
    if (res.a[10] > y.a[i]) res.add(y.a[i]);
  return res;
}
void dfs(int u, int tr) {
  for (long long i = (0); i <= (int(ke[u].size()) - 1); i++) {
    int v = ke[u][i];
    if (v == tr) continue;
    h[v] = h[u] + 1;
    par[v][0] = u;
    for (long long j = (1); j <= (19); j++) {
      par[v][j] = par[par[v][j - 1]][j - 1];
      gt[v][j] = addnode(gt[v][j - 1], gt[par[v][j - 1]][j - 1]);
    }
    dfs(v, u);
  }
}
int lca(int u, int v) {
  if (h[u] < h[v]) swap(u, v);
  int dif = h[u] - h[v];
  for (long long i = (19); i >= (0); i--)
    if ((dif >> i) & 1) u = par[u][i];
  if (u == v) return u;
  for (long long i = (19); i >= (0); i--)
    if (par[u][i] != par[v][i]) {
      u = par[u][i];
      v = par[v][i];
    }
  return par[u][0];
}
node getup(int v, int dif) {
  node res;
  res.reset();
  for (long long i = (19); i >= (0); i--)
    if ((dif >> i) & 1) {
      res = addnode(res, gt[v][i]);
      v = par[v][i];
    }
  return res;
}
void setup() {
  cin >> n >> m >> q;
  for (long long i = (1); i <= (n); i++)
    for (long long j = (0); j <= (19); j++) gt[i][j].reset();
  for (long long i = (1); i <= (n - 1); i++) {
    int u, v;
    cin >> u >> v;
    ke[u].push_back(v);
    ke[v].push_back(u);
  }
  for (long long i = (1); i <= (m); i++) {
    int x;
    cin >> x;
    gt[x][0].add(i);
  }
}
void work() {
  dfs(1, 0);
  while (q--) {
    int u, v, a;
    cin >> u >> v >> a;
    int l = lca(u, v);
    node x = getup(u, h[u] - h[l]);
    node y = getup(v, h[v] - h[l] + 1);
    node ans = addnode(x, y), res;
    res.reset();
    if (giong(ans, res)) {
      cout << "0\n";
      continue;
    }
    int k = 10;
    while (ans.a[k] > m && k > -1) k--;
    if (k == 10) k--;
    cout << min(a, k + 1) << " ";
    for (long long i = (0); i <= (min(a - 1, k)); i++) cout << ans.a[i] << " ";
    cout << "\n";
  }
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie();
  cout.tie();
  setup();
  work();
  return 0;
}